Can someone please help me out?
Calculate the above formula, expressing the answer in scientific notation with the correct number of significant figures:
hint: \[\Large 1 \times {10^{ - 3}} = 0.001\]
and: \[\Large 3.61 \times {10^{ - 3}} = 0.00361\]
from your attachement I see this computation: \[\Large \left( {8.86 + 1.0 \times {{10}^{ - 3}}} \right) \div \left( {3.61 \times {{10}^{ - 3}}} \right)\] am I right?
Yea you are right. I accidentally read the sign wrong. I was suppose to divide not add. So I should get 2.45429362881 right?
I got: \[\Large \left( {8.86 + 1.0 \times {{10}^{ - 3}}} \right) \div \left( {3.61 \times {{10}^{ - 3}}} \right) = 2.454 \times {10^3}\]
Do I need the full number or should I just shorten it to 2.454?
I think that we have to take at least 3 significant figures, so our result will be like below: \[\Large 2.45 \times {10^3}\]
What exactly are the significant figures? I was confused on that part.
the significant figures are 2, 4, and 5
another example: if we have a number like below: 1.0025 then the significant figures are: 1, 2, and 5
whereas, if we have this number: 1,200 then the significant figures are 1 and 2
So the signification figures would the answer to that part be 3? Or would the answer to that part be 2, 4, and 5.
if we have this number: 0.0025 then the significant figures are 2, and 5
your answer, is: \[\Large 2.45 \times {10^3}\]
So with scientific notation you only go 2 digits after the decimal right? I thought the answer was 2.454 x 10^-3. Or am I wrong on that part?
when we have a decimal point, the significant figures are the figures at the left and at the right with respect to that decimal point
so, the number 2.454 x 10^-3 has four significant figures
please, wait, is your computation an experimental measure?
I'm not sure what you mean on computation or experimental?
since 1*10^-3 has only one significant figure, then we have to rewrite our result with one significant figure, so your result is: \[\Large 2 \times {10^3}\]
I was just wondering on the scientific noation is both 2.45 and 2.454 both correct. It doesn't matter how I state it. Only difference is how many significant figures their are.
2.45 has three significant figures, whereas 2.454 has four significant figures
as a general prescription, we have to round off our result like the most inaccurate measure. Now our most inaccurate number is 1*10^-3 which has only one significant figure. So we have to round off our final result to one significant figure, so we get: \[\Large 2 \times {10^3}\]
I'm confused on how I'm suppose to state my answer. How should I write it?
it is simple, your measure has to be written like below: \[\Large 2 \times {10^3}\] which has only one significan figure, and such significant figure is 2
I have rounded off 2.454... to 2
please sorry, I see that the most inaccurate number is: 1.0 *10^-3 which has 2 significant figures, so we have to round off our result to 2 significant figures, so your result is: \[\Large 2.5 \times {10^3}\] I have rounded off 2.454... to 2.5 and 2.5 *10^3 has 2 significant figures, namely 2, and 5 Sorry again, for my error
more explanation: a number like this one: 1.0 has two significant figures, namely 1 and 0
Do I have to round when stating my answer? Does it matter how I state my answer also?
yes! since when you do a computation, using some formula, the result which you get, can not be as accurate as the most accurate measure
as a general prescription, the result which you get from a computation has to be as accurate as the less accurate measure
So how many digits should I state in the answer? Should I round it or not? I normally don't round it unless the question states to do so.
yes! you have to round it, like below: \[\Large 2.5 \times {10^3}\]
So is the answer is 2.5 x 10^3 or 2.5 x 10^-3 @Michele_Laino ? Do you need to take the - off with scientific notation? The original question had a -3 so I wasn't sure. Also for the significant figures there's two of them right? 2 and 5.
yes! the significant figures are 2 and 5, and the right result is 2.5 x 10^3
Thanks for all your help @Michele_Laino
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